LeetCode96_Unique Binary Search Trees(求1到n这些节点能够组成多少种不同的二叉查找树) Java题解
2017-06-03 09:24
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题目:
Given n, how many structurally unique BST's (binary search trees) that store values 1...n?
For example,
Given n = 3, there are a total of 5 unique BST's.
1 3 3 2 1 \ / / / \ \ 3 2 1 1 3 2 / / \ \ 2 1 2 3解题:
用递归的思想,当仅仅有0个或是1个节点的时候。仅仅有一种。n个节点的时候有f(n)种:
左边能够有n-1个节点,右边0个节点,依据对称性能够左右互换。这时候有2*f(n-1)*f(0);
一边1个,还有一边n-2个,这时候有2*f(1)*f(n-2);
一边两个,一边N-3个,这时候有2*f(2)*f(n-3);
。。。
。。。
假设n为奇数,两边都是n/2个。这时候有f(n/2)*f(n/2),假设n为偶数,一边n/2一边(n/2+1)个,为2*f(n/2)*f(n/2+1);
代码:
public static int numTrees(int n) { if(n==1||n==0) return 1; int sum=0; if(n%2==0) { for(int k=n-1;k>=n/2;k--) { sum+=2*numTrees(k)*numTrees(n-1-k); } return sum; } else { for(int k=n-1;k>n/2;k--) { sum+=2*numTrees(k)*numTrees(n-1-k); } return sum+numTrees(n/2)*numTrees(n/2); } }
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